// Workers AI · dad joke modeWhy was the special trinomial sad? It was a cubic failure.

In algebra, specifically in the factorization of polynomials, a special trinomial is a polynomial that can be expressed in the form:

$ax^{2}+bx+c\qquad ({\mbox{with }}a,b,c\neq 0)$

and for which there is a known method for factoring it as the product of two first-degree binomials.

Factoring Method

We can distinguish between the cases in which the coefficient of the quadratic term is equal to or different from $1$.

a = 1

If the coefficient of the quadratic term is equal to $1$, the trinomial takes the form:

$x^{2}+bx+c\,$

in this case, it can be factored into the product of two first-degree binomials, in the form:

$(x+u)(x+v)$,

where $u$ and $v$ are two terms with the following two properties:

• $u+v=b$
• $u\cdot v=c$.

In fact, performing the calculations yields:

${\begin{aligned}(x+u)(x+v)&=x^{2}+vx+ux+uv\\&=x^{2}+x(v+u)+uv\\&=x^{2}+bx+c\\\end{aligned}}$

A practical method to find $u$ and $v$ is to solve for the two roots of the polynomial. In fact, if

$x^{2}+bx+c=0$,

then:

$(x-u)(x-v)=0\Rightarrow {\begin{cases}x_{1}=u\\x_{2}=v\end{cases}}$

To find the roots of the quadratic trinomial, simply use the quadratic formula:

$x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

a ≠ 1

If the coefficient of the quadratic term is not $1$, the trinomial can be factored as follows:

$ax^{2}+bx+c=a(x+u)(x+v)$,

where $u$ and $v$ have the following properties:

• $u+v={\frac {b}{a}}$
• $u\cdot v={\frac {c}{a}}$.

In this case as well, the factorization can be demonstrated as follows:

${\begin{aligned}a(x+u)(x+v)&=ax^{2}+avx+aux+auv\\&=ax^{2}+ax(v+u)+auv\\&=ax^{2}+bx+c\\\end{aligned}}$

As in the previous case, $u$ and $v$ can be found by solving for the roots of the polynomial using the quadratic formula.

Trinomials of degree greater than 2

Factoring by substitution

More generally, if we consider the trinomial:

$ax^{2n}+bx^{n}+c$

This can be factored by making the substitution $x^{n}=t$, which gives the trinomial:

$at^{2}+bt+c$,

that can be factored using the methods described above and then by reapplying the substitution in reverse.

Direct factoring

Considering the previous properties:

• $u+v={\frac {b}{a}}$
• $u\cdot v={\frac {c}{a}}$.

The trinomial can also be directly factored as follows:

$ax^{2n}+bx^{n}+c=a(x^{n}+u)(x^{n}+v)$

The factorization can be demonstrated as follows:

${\begin{aligned}a(x^{n}+u)(x^{n}+v)&=ax^{2n}+ax^{n}u+ax^{n}v+auv\\&=ax^{2n}+ax^{n}(u+v)+auv\\&=ax^{2n}+ax^{n}{\frac {b}{a}}+a{\frac {c}{a}}\\&=ax^{2n}+bx^{n}+c\end{aligned}}$

Also in this case $u$ and $v$ can be found by solving for the roots of the polynomial using the quadratic formula.

Bibliography

• Marzia Re Fraschini, Gabriella Grazzi (2012). I principi della matematica (Volume 3). Atlas. ISBN 978-88-268-1711-8.
• Massimo Bergamini, Graziella Barozzi, Anna Trifone. Matematica.blu (seconda edizione) Vol.1. Zanichelli - Bologna, 2018. ISBN 978-88-08-22085-1.{{cite book}}: CS1 maint: multiple names: authors list (link)